Question

1. let $f(x) = \sqrt{x + 1}$, $g(x) = 2x - 5$, and $h(x) = 3x^2 - 3$. find the indicated value. $h(f(10)) = \square$

Explanation:

Step1: Calculate \( f(10) \)

Given \( f(x)=\sqrt{x + 1} \), substitute \( x = 10 \):
\( f(10)=\sqrt{10 + 1}=\sqrt{11} \)? Wait, no, wait, \( 10 + 1 = 11 \)? Wait, no, wait, the problem is \( f(x)=\sqrt{x + 1} \), so \( f(10)=\sqrt{10 + 1}=\sqrt{11} \)? Wait, no, wait, maybe I made a mistake. Wait, no, let's check again. Wait, the function \( f(x)=\sqrt{x + 1} \), so when \( x = 10 \), \( f(10)=\sqrt{10 + 1}=\sqrt{11} \)? Wait, no, wait, maybe the original problem has a typo? Wait, no, wait, the user's problem: \( f(x)=\sqrt{x + 1} \), \( g(x)=2x - 5 \), \( h(x)=3x^2 - 3 \). Wait, but the question is \( h(f(10)) \). So first, find \( f(10) \), then substitute into \( h(x) \).

Wait, \( f(10)=\sqrt{10 + 1}=\sqrt{11} \)? Wait, no, wait, \( 10 + 1 = 11 \), so \( f(10)=\sqrt{11} \). Then \( h(f(10)) = h(\sqrt{11}) = 3(\sqrt{11})^2 - 3 \). Since \( (\sqrt{11})^2 = 11 \), so \( 3\times11 - 3 = 33 - 3 = 30 \). Wait, that makes sense. Let's redo:

Step1: Compute \( f(10) \)

\( f(x)=\sqrt{x + 1} \), so \( f(10)=\sqrt{10 + 1}=\sqrt{11} \)? Wait, no, wait, \( 10 + 1 = 11 \), so \( f(10)=\sqrt{11} \).

Step2: Compute \( h(f(10)) \)

\( h(x)=3x^2 - 3 \), substitute \( x = f(10)=\sqrt{11} \):
\( h(\sqrt{11}) = 3\times(\sqrt{11})^2 - 3 \)
Since \( (\sqrt{11})^2 = 11 \), then \( 3\times11 - 3 = 33 - 3 = 30 \).

Wait, but maybe I misread the function. Wait, maybe \( f(x)=\sqrt{x} + 1 \)? Let me check the original problem again. The user's image: "Let \( f(x)=\sqrt{x + 1} \), \( g(x)=2x - 5 \), \( h(x)=3x^2 - 3 \). Find the indicated value \( h(f(10)) \)."

So yes, \( f(10)=\sqrt{10 + 1}=\sqrt{11} \), then \( h(\sqrt{11}) = 3(\sqrt{11})^2 - 3 = 3\times11 - 3 = 30 \).

Wait, but maybe the function is \( f(x)=\sqrt{x} + 1 \)? Let's see, if \( f(x)=\sqrt{x} + 1 \), then \( f(10)=\sqrt{10} + 1 \), then \( h(f(10)) = 3(\sqrt{10} + 1)^2 - 3 \), which is more complicated. But the original problem says \( f(x)=\sqrt{x + 1} \), so we go with that.

So:

Step1: Calculate \( f(10) \)

\( f(10) = \sqrt{10 + 1} = \sqrt{11} \)

Step2: Calculate \( h(f(10)) \)

\( h(x) = 3x^2 - 3 \), substitute \( x = \sqrt{11} \):
\( h(\sqrt{11}) = 3(\sqrt{11})^2 - 3 = 3 \times 11 - 3 = 33 - 3 = 30 \)

Answer:

\( 30 \)